Optics - Refraction and a transparent sphere

In summary: I think that the angle you have labelled as 60° is 120° because the sum of the angles of both drawings is 180°. I also think that the ray from the spider meets the circle again at the point where the insect was initially placed.
  • #1
Saitama
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Optics -- Refraction and a transparent sphere

Homework Statement


A spider is hanging by means of its own silk thread directly above a transparent fixed sphere of radius R=20 cm as shown in the figure. The refractive index of the material of the sphere is equal to ##\sqrt{2}## and the height of the spider from the centre of the sphere is 2R. An insect, initially sitting at the bottom, starts crawling on the sphere along a vertical circular path with the constant speed of ##\frac{\pi}{4}## cm/s. For how long time the insect will be invisible for the spider, assume that it crawls once round the vertical circle.
(see attachment)

Homework Equations





The Attempt at a Solution


Honestly, I have no idea. I don't have a clue about which equations I have to use here.

Any help is appreciated. Thanks!
 

Attachments

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  • #2


Pranav-Arora said:

Homework Statement


A spider is hanging by means of its own silk thread directly above a transparent fixed sphere of radius R=20 cm as shown in the figure. The refractive index of the material of the sphere is equal to ##\sqrt{2}## and the height of the spider from the centre of the sphere is 2R. An insect, initially sitting at the bottom, starts crawling on the sphere along a vertical circular path with the constant speed of ##\frac{\pi}{4}## cm/s. For how long time the insect will be invisible for the spider, assume that it crawls once round the vertical circle.
(see attachment)

Homework Equations



The Attempt at a Solution


Honestly, I have no idea. I don't have a clue about which equations I have to use here.

Any help is appreciated. Thanks!
This question has to do with total internal reflection. Start there.
 
  • #3


SammyS said:
This question has to do with total internal reflection. Start there.

I did think of this before but the insect keeps on moving which is confusing me.
 
  • #4


Pranav-Arora said:
I did think of this before but the insect keeps on moving which is confusing me.
Find out the location of the insect just as a ray from the insect to the spider reaches the point of criticality.

I actually found this easier to do by tracing the ray from the spider, back to the insect.
 
  • #5


SammyS said:
I actually found this easier to do by tracing the ray from the spider, back to the insect.

I too think this would be easier.
If a ray from the spider is tangent to the sphere, then by Snell's law, it deviates towards the normal by pi/4. (see attachment)

I am facing problems with geometry. Look at the triangle where the angles come out to be 150 and 45 degrees, the sum goes over 180 degrees! I can't find where I went wrong.

EDIT: Woops, please ignore this reply. I think I have found out my mistake. One more question, would the rays meet, after refraction, at the point where insect was initially placed?
 

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  • #6


Pranav-Arora said:
I too think this would be easier.
If a ray from the spider is tangent to the sphere, then by Snell's law, it deviates towards the normal by pi/4. (see attachment)

I am facing problems with geometry. Look at the triangle where the angles come out to be 150 and 45 degrees, the sum goes over 180 degrees! I can't find where I went wrong.

EDIT: Woops, please ignore this reply. I think I have found out my mistake. One more question, would the rays meet, after refraction, at the point where insect was initially placed?
They don't meet there, if my solution is correct.

attachment.php?attachmentid=55855&d=1361161755.png


The angle you have labelled as 60° is 120° .
 
  • #7


SammyS said:
The angle you have labelled as 60° is 120° .

Yes, I figured that out already, that's why I said to ignore the reply. I will be back with a solution. :smile:
 
  • #8


Pranav-Arora said:
Yes, I figured that out already, that's why I said to ignore the reply. I will be back with a solution. :smile:

After being refracted, the ray from the spider then intercepts the circle again. A radius from each place the ray intercepts the circle together with the ray itself, form an isosceles right triangle, with the ray falling along the hypotenuse.
 
  • #9


I am still not getting the right answer.
(see attachment)
The angular displacement of the insect when it is invisible to the spider is ##2\cdot \frac{13\pi}{24}##.
The time taken is ##t=\frac{\theta}{\omega}## where ##\omega## is the angular velocity of insect.
Substituting the values, ##t=\frac{260}{3} sec## which is wrong. :(
 

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  • #10


Your drawings are misleading.

The refracted ray makes an angle of 45° (π/2) with respect to the normal .

In both drawings, the refracted ray appears to be drawn at an angle of 45° with respect to the horizontal.
 
  • #11


ehild said:
The spider is visible while it crawls along the red arc. What central angle belongs to it?

ehild

How do you know that the rays go like you have shown in your diagram? :confused:
 
  • #12


The bug is visible along the red arc. The refractive index is 1/√2 so the angle of refraction is 45°. The radii make 90°angle between the point of incidence and point where the refracted ray arrives to the circle again.

ehild
 

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  • #13


Thank you both for your inputs, I have got the right answer! :smile:
 
  • #14


Pranav-Arora said:
How do you know that the rays go like you have shown in your diagram? :confused:
You, yourself, said the angle of refraction is π/4 .
 
  • #15


SammyS said:
You, yourself, said the angle of refraction is π/4 .

Yes, I did but I was confused by the path the rays would take after refraction. I messed up with simple geometry. :redface:
 
  • #16


Pranav-Arora said:
Yes, I did but I was confused by the path the rays would take after refraction. I messed up with simple geometry. :redface:

The rays travel along straight lines :tongue2:
 
  • #17


ehild said:
The rays travel along straight lines :tongue2:

:biggrin:
Yes but see my previous diagrams, they are completely wrong.
 
  • #18


Yes, the upper central angle was wrong, and also, you did not draw the refracted light according to the angle you got.

ehild.
 

1. What is refraction?

Refraction is the bending of light as it passes through a medium, such as air, water, or a transparent sphere. This bending is caused by a change in the speed of light when it travels from one medium to another.

2. How does a transparent sphere affect the refraction of light?

A transparent sphere can act as a lens, causing light to bend and converge at a focal point. The amount of refraction depends on the curvature and index of refraction of the sphere.

3. What is the index of refraction?

The index of refraction is a measure of how much a medium can bend light. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. The higher the index of refraction, the more a medium can bend light.

4. How does the angle of incidence affect refraction?

The angle of incidence, or the angle at which light enters a medium, is directly related to the angle of refraction. As the angle of incidence increases, the angle of refraction also increases, causing the light to bend more.

5. What is Snell's Law?

Snell's Law, also known as the law of refraction, describes the relationship between the angle of incidence and the angle of refraction. It states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the indices of refraction for the two mediums.

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